Does this equation have a known solution given a starting value $\phi_1$?

$$\phi_{n}\prod_{j=1}^{n-1}(1-\phi_{j}^2) + \phi_{n-1}\phi_1=0$$

Background: I am trying to derive the partial autocorrelation function of a first-order moving average process $Y_t=\epsilon_t+\theta\epsilon_{t-1}$. There,


and the goal is to show that

$$ \phi_n=\frac{-(-\theta)^n}{1+\theta^2+\cdots+\theta^{2n}} $$

See e.g. my partial answer here.


@Elaqqad, I cannot replicate that the formula is wrong for $n=4$ (or $n=5$ for that matter):

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  • $\begingroup$ the given "goal" is not true:$\phi_2=\frac{-\phi_1^2}{1-\phi_1^2}$ which gives $\frac{\theta^2}{1+2\theta^2}$ $\endgroup$
    – Elaqqad
    Mar 5, 2015 at 13:43
  • $\begingroup$ $-\left(\frac{\theta}{1+\theta^2}\right)^2\Big/\left(1-\left(\frac{\theta}{1 + \theta^2}\right)^2\right)=-\theta^2/[1+2\theta^2+\theta^4-\theta^2]=-\theta^2/[1 + \theta^2 +\theta^4]$ ? $\endgroup$ Mar 5, 2015 at 14:07
  • $\begingroup$ my mistake, but I think it's not the right formula $\endgroup$
    – Elaqqad
    Mar 5, 2015 at 14:15
  • $\begingroup$ which one? the goal, or the first display? $\endgroup$ Mar 5, 2015 at 14:22
  • $\begingroup$ For me the first display it's an equation (true for me) and we want a closed form for $\phi_n$ so when I try to inject your formula in the equation it does not give what we want, finally:$$\phi_3=\frac{-\theta^3}{(1+\theta^2+\theta^4+\theta^6)} $$ check it if i did not made any mistake, so your formula works ! $\endgroup$
    – Elaqqad
    Mar 5, 2015 at 14:34

1 Answer 1


There is another recurrence relation for your $\phi_n$, to prove it let us first return to your recurrence relation: $$\phi_{n} =\frac{-\phi_{n-1}\phi_1}{\prod_{j=1}^{n-1}(1-\phi_{j}^2)}$$

So if we tray to express $\phi_{n+1}$ using $\phi_n$ we can get the following results:

$$\begin{align}\phi_{n+1} &=&&\frac{-\phi_{n}\phi_1}{\prod_{j=1}^{n}(1-\phi_{j}^2)}\\ &=&&\frac{\phi_n}{\phi_{n-1}(1-\phi_n^2)}. \frac{-\phi_{n-1}\phi_1}{\prod_{j=1}^{n-1}(1-\phi_{j}^2)}\\ &=&&\frac{\phi_n^2}{\phi_{n-1}(1-\phi_n^2)} \end{align}$$

So the sequence can be defined using the following relations: $$\begin{align} \phi_1&=&&\frac{\theta}{1+\theta^2}\\ \phi_2&=&&\frac{-\theta^2}{1+\theta^2+\theta^4}\\ \phi_{n+1}&=&&\frac{\phi_n^2}{\phi_{n-1}(1-\phi_n^2)} \end{align} $$ We consider $f(n)=1+\theta^2+\cdots+\theta^{2n}$. so let's prove by induction that:

$$ \phi_n=\frac{-(-\theta)^n}{f(n)} \ \ \ (*) $$

first $(*)$ is true for $n=0,1,2...$, so assume that $(*)$ is true for $n-1$ and $n$ we have : $$\begin{align} \phi_{n+1}&=&&\frac{\phi_n^2}{\phi_{n-1}(1-\phi_n^2)}\\ &=&&\frac{\theta^{2n}f(n-1)}{f(n)^2.(-(-\theta)^{n-1}) (1-\frac{\theta^{2n}}{f(n)^2})}\\ &=&&-(-\theta)^{n+1}\frac{f(n-1)}{f(n)^2-\theta^{2n}} \end{align}$$ because $-(-\theta)^{n+1}.-(-\theta)^{n-1}=\theta^{2n}$ and we have also : $$\begin{align}f(n)^2-\theta^{2n}&=&&f(n-1)+\theta^{2n})^2-\theta^{2n}\\ &=&&f(n-1)^2+2\theta^{2n}f(n-1)+\theta^{2n}(\theta^{2n}-1)\\ &=&&f(n-1)^2+\theta^{2n}f(n-1)+\theta^{2n}(\theta^2-1)f(n-1)\\ &=&&f(n-1)(f(n-1)+2\theta^{2n}+\theta^{2(n+1)}-\theta^{2n})\\ &=&&f(n-1)f(n+1) \end{align}$$

finally : $$\phi_{n+1}=\frac{-(-\theta)^{n+1}}{f(n+1)} $$

  • $\begingroup$ I'm sorry for the first incorrect answer, check this proof and let me know $\endgroup$
    – Elaqqad
    Mar 5, 2015 at 16:28
  • $\begingroup$ Nice, thanks a lot! I updated my answer on Cross Validated to link to your derivation. $\endgroup$ Mar 6, 2015 at 7:26

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